Bài toán 2: $ \cos \left[ {\frac{\pi }{2} - \pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$ \Leftrightarrow \sin \left[ {\pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$
\Leftrightarrow \left[ {\begin{matrix} \pi \left( x^2 + 2x \right) =
\pi x^2 + k2 \pi \\ \pi (x^2 + 2x) = \pi - \pi x^2 + k2\pi
\end{matrix}} \right. $$ \Leftrightarrow \left[ {\begin{matrix} x = k
\in \mathbb{Z} \\ 2x^2 + 2x - \left( 2k + 1 \right) = 0 \end{matrix}}
\right. $$\left( {\text{*}} \right)$Do $\begin{cases}\left(
{\text{*}} \right) \\x{\text{ > }}0 \\k \in \mathbb{Z}
\\\end{cases} $ suy ra $\min x = \frac{{\sqrt 3 - 1}}{2}$
Bài toán 2: $ \cos \left[ {\frac{\pi }{2} - \pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$ \Leftrightarrow \sin \left[ {\pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$
\Leftrightarrow \left[ {\begin{matrix} \pi \left( x^2 + 2x \right) =
\pi x^2 + k2 \pi \\ \pi x^2 + 2x = \pi - \pi x^2 + k2\pi
\end{matrix}} \right. $$ \Leftrightarrow \left[ {\begin{matrix} x = k
\in \mathbb{Z} \\ 2x^2 + 2x - \left( 2k + 1 \right) = 0 \end{matrix}}
\right. $$\left( {\text{*}} \right)$Do $\begin{cases}\left(
{\text{*}} \right) \\x{\text{ > }}0 \\k \in \mathbb{Z}
\\\end{cases} $ suy ra $\min x = \frac{{\sqrt 3 - 1}}{2}$
Bài toán 2: $ \cos \left[ {\frac{\pi }{2} - \pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$ \Leftrightarrow \sin \left[ {\pi \left( {{x^2} + 2x} \right)} \right] = \sin \left( {\pi {x^2}} \right)$$
\Leftrightarrow \left[ {\begin{matrix} \pi \left( x^2 + 2x \right) =
\pi x^2 + k2 \pi \\ \pi
(x^2 + 2x
) = \pi - \pi x^2 + k2\pi
\end{matrix}} \right. $$ \Leftrightarrow \left[ {\begin{matrix} x = k
\in \mathbb{Z} \\ 2x^2 + 2x - \left( 2k + 1 \right) = 0 \end{matrix}}
\right. $$\left( {\text{*}} \right)$Do $\begin{cases}\left(
{\text{*}} \right) \\x{\text{ > }}0 \\k \in \mathbb{Z}
\\\end{cases} $ suy ra $\min x = \frac{{\sqrt 3 - 1}}{2}$